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using a 0 to indicate the left branch and a 1 to indicate the right branch For instance, s is reached by going left, then right, and finally right This is encoded as 0 1 1 If character c is at depth d i and occurs f; times, the cost of the code is C dif; We can obtain a better code than the one given in Figure 132 by recognizing that nl is an only child By placing it one level higher (replacing its parent), we obtain the new tree shown in Figure 133 This new tree has a cost of 173 but is still far from optimal Note that the tree in Figure 133 is a full tree, in which all nodes either are leaves or have two children An optimal code always has this property; otherwise, as already shown, nodes with only one child could move up a level If the characters are placed only at the leaves, any sequence of bits can always be decoded unambiguously For instance, suppose that the encoded string is 0 1001 1 1 1000 10 1 1000 10001 11 Figure 133 shows that 0 and 01 are not character codes but that 010 represents i, so the first character is i Then 01 1 follows, which is an s Then 11 follows, which is a newline (nl) The remainder of the code is a, sp, t, i, e, and nl The character codes can be different lengths, so long as no character code is a prefix of another character code, an encoding called a prefix code Conversely, if a character is contained in a nonleaf node, guaranteeing unambiguous decoding is no longer possible Thus our basic problem is to find the full binary tree of minimum cost (as defined previously) in which all characters are contained in the leaves The tree shown in Figure 134 is optimal for our sample alphabet As shown.

In a prefix code, no character is a prefix of another character codeThis is guaranteed if the Characters are only in leaves A prefix code can be decoded unambiguously

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The maximum likelihood estimates of f3 and 0' are chosen to maximize the likelihood function If the family of possible error distributions is the family of normal distributions with mean 0, then the ML estimates of the regression coefficients coincide with the least-squares estimates; the ML estimate of 0' is (.; (n - p - 1) In )cT LS For the family of Laplace (double exponential) distributions with mean 0, the ML estimates of the regression coefficients coincide with the LAD estimates..

in Figure 135, this code requires only 146 bits There are many optimal codes, which can be obtained by swapping children in the encoding tree

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HOWis the coding tree constructed The coding system algorithm was given by Huffman in 1952 Commonly called Huffman's algorithm, it constructs an optimal prefix code by repeatedly merging trees until the final tree is obtained Throughout this section, the number of characters is C In Huffman's algorithm we maintain a forest of trees The weight of a tree is the sum of the frequencies of its leaves C - 1 times, two trees, T , and T 2 ,of smallest weight are selected, breaking ties arbitrarily, and a new tree is formed with subtrees T , and T Z At the beginning of the algorithm, there are C single-node trees (one

The cumulative evidence of many experiments is that indeed only PR (and PL ) are involved in weak interactions. The absence of the "mirror image" states, PL and P R , is a clear violation of parity invariance (see Section 5.7). Also, charge conjugation, C, invariance is violated, since C transforms a PL state into a Pl. state. However, the y"(l - y5) form leaves the weak interaction invariant under the combined CP operation. For instance,

for each character) At the end of the algorithm, there is one tree, giving an optimal Huffman tree In Exercise 134 you are asked to prove Huffman's algorithm gives an optimal tree An example helps make operation of the algorithm clear Figure 136 shows the initial forest; the weight of each tree is shown in small type at the root The two trees of lowest weight are merged, creating the forest shown in Figure 137 The new root is T I We made s the left child arbitrarily; any tiebreaking procedure can be used The total weight of the new tree is just the sum of the weights of the old trees and can thus be easily computed Now there are six trees, and we again select the two trees of smallest weight, T1 and t They are merged into a new tree with root T 2 and weight 8, as shown in Figure 138 The third step merges T 2 and a, creating T3, with weight 10 + 8 = 18 Figure 139 shows the result of this operation

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